3.200 \(\int (d+e x) \log (c (a+\frac{b}{x})^p) \, dx\)

Optimal. Leaf size=78 \[ -\frac{p (a d-b e)^2 \log (a x+b)}{2 a^2 e}+\frac{(d+e x)^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e}+\frac{b e p x}{2 a}+\frac{d^2 p \log (x)}{2 e} \]

[Out]

(b*e*p*x)/(2*a) + ((d + e*x)^2*Log[c*(a + b/x)^p])/(2*e) + (d^2*p*Log[x])/(2*e) - ((a*d - b*e)^2*p*Log[b + a*x
])/(2*a^2*e)

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Rubi [A]  time = 0.0561594, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {2463, 514, 72} \[ -\frac{p (a d-b e)^2 \log (a x+b)}{2 a^2 e}+\frac{(d+e x)^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e}+\frac{b e p x}{2 a}+\frac{d^2 p \log (x)}{2 e} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*Log[c*(a + b/x)^p],x]

[Out]

(b*e*p*x)/(2*a) + ((d + e*x)^2*Log[c*(a + b/x)^p])/(2*e) + (d^2*p*Log[x])/(2*e) - ((a*d - b*e)^2*p*Log[b + a*x
])/(2*a^2*e)

Rule 2463

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))*((f_.) + (g_.)*(x_))^(r_.), x_Symbol] :> Simp[((
f + g*x)^(r + 1)*(a + b*Log[c*(d + e*x^n)^p]))/(g*(r + 1)), x] - Dist[(b*e*n*p)/(g*(r + 1)), Int[(x^(n - 1)*(f
 + g*x)^(r + 1))/(d + e*x^n), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, p, r}, x] && (IGtQ[r, 0] || RationalQ[n
]) && NeQ[r, -1]

Rule 514

Int[(x_)^(m_.)*((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[x^(m - n*q)*
(a + b*x^n)^p*(d + c*x^n)^q, x] /; FreeQ[{a, b, c, d, m, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] |
|  !IntegerQ[p])

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rubi steps

\begin{align*} \int (d+e x) \log \left (c \left (a+\frac{b}{x}\right )^p\right ) \, dx &=\frac{(d+e x)^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e}+\frac{(b p) \int \frac{(d+e x)^2}{\left (a+\frac{b}{x}\right ) x^2} \, dx}{2 e}\\ &=\frac{(d+e x)^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e}+\frac{(b p) \int \frac{(d+e x)^2}{x (b+a x)} \, dx}{2 e}\\ &=\frac{(d+e x)^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e}+\frac{(b p) \int \left (\frac{e^2}{a}+\frac{d^2}{b x}-\frac{(a d-b e)^2}{a b (b+a x)}\right ) \, dx}{2 e}\\ &=\frac{b e p x}{2 a}+\frac{(d+e x)^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )}{2 e}+\frac{d^2 p \log (x)}{2 e}-\frac{(a d-b e)^2 p \log (b+a x)}{2 a^2 e}\\ \end{align*}

Mathematica [A]  time = 0.0269028, size = 85, normalized size = 1.09 \[ \frac{1}{2} b e p \left (\frac{x}{a}-\frac{b \log (a x+b)}{a^2}\right )+d x \log \left (c \left (a+\frac{b}{x}\right )^p\right )+\frac{1}{2} e x^2 \log \left (c \left (a+\frac{b}{x}\right )^p\right )+\frac{b d p \log \left (a+\frac{b}{x}\right )}{a}+\frac{b d p \log (x)}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*Log[c*(a + b/x)^p],x]

[Out]

(b*d*p*Log[a + b/x])/a + d*x*Log[c*(a + b/x)^p] + (e*x^2*Log[c*(a + b/x)^p])/2 + (b*d*p*Log[x])/a + (b*e*p*(x/
a - (b*Log[b + a*x])/a^2))/2

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) \ln \left ( c \left ( a+{\frac{b}{x}} \right ) ^{p} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)*ln(c*(a+b/x)^p),x)

[Out]

int((e*x+d)*ln(c*(a+b/x)^p),x)

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Maxima [A]  time = 1.02641, size = 74, normalized size = 0.95 \begin{align*} \frac{1}{2} \, b p{\left (\frac{e x}{a} + \frac{{\left (2 \, a d - b e\right )} \log \left (a x + b\right )}{a^{2}}\right )} + \frac{1}{2} \,{\left (e x^{2} + 2 \, d x\right )} \log \left ({\left (a + \frac{b}{x}\right )}^{p} c\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*log(c*(a+b/x)^p),x, algorithm="maxima")

[Out]

1/2*b*p*(e*x/a + (2*a*d - b*e)*log(a*x + b)/a^2) + 1/2*(e*x^2 + 2*d*x)*log((a + b/x)^p*c)

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Fricas [A]  time = 1.68494, size = 184, normalized size = 2.36 \begin{align*} \frac{a b e p x +{\left (2 \, a b d - b^{2} e\right )} p \log \left (a x + b\right ) +{\left (a^{2} e x^{2} + 2 \, a^{2} d x\right )} \log \left (c\right ) +{\left (a^{2} e p x^{2} + 2 \, a^{2} d p x\right )} \log \left (\frac{a x + b}{x}\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*log(c*(a+b/x)^p),x, algorithm="fricas")

[Out]

1/2*(a*b*e*p*x + (2*a*b*d - b^2*e)*p*log(a*x + b) + (a^2*e*x^2 + 2*a^2*d*x)*log(c) + (a^2*e*p*x^2 + 2*a^2*d*p*
x)*log((a*x + b)/x))/a^2

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Sympy [A]  time = 3.9264, size = 156, normalized size = 2. \begin{align*} \begin{cases} d p x \log{\left (a + \frac{b}{x} \right )} + d x \log{\left (c \right )} + \frac{e p x^{2} \log{\left (a + \frac{b}{x} \right )}}{2} + \frac{e x^{2} \log{\left (c \right )}}{2} + \frac{b d p \log{\left (x + \frac{b}{a} \right )}}{a} + \frac{b e p x}{2 a} - \frac{b^{2} e p \log{\left (x + \frac{b}{a} \right )}}{2 a^{2}} & \text{for}\: a \neq 0 \\d p x \log{\left (b \right )} - d p x \log{\left (x \right )} + d p x + d x \log{\left (c \right )} + \frac{e p x^{2} \log{\left (b \right )}}{2} - \frac{e p x^{2} \log{\left (x \right )}}{2} + \frac{e p x^{2}}{4} + \frac{e x^{2} \log{\left (c \right )}}{2} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*ln(c*(a+b/x)**p),x)

[Out]

Piecewise((d*p*x*log(a + b/x) + d*x*log(c) + e*p*x**2*log(a + b/x)/2 + e*x**2*log(c)/2 + b*d*p*log(x + b/a)/a
+ b*e*p*x/(2*a) - b**2*e*p*log(x + b/a)/(2*a**2), Ne(a, 0)), (d*p*x*log(b) - d*p*x*log(x) + d*p*x + d*x*log(c)
 + e*p*x**2*log(b)/2 - e*p*x**2*log(x)/2 + e*p*x**2/4 + e*x**2*log(c)/2, True))

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Giac [A]  time = 1.26641, size = 151, normalized size = 1.94 \begin{align*} \frac{a^{2} p x^{2} e \log \left (a x + b\right ) - a^{2} p x^{2} e \log \left (x\right ) + 2 \, a^{2} d p x \log \left (a x + b\right ) + a^{2} x^{2} e \log \left (c\right ) - 2 \, a^{2} d p x \log \left (x\right ) + a b p x e + 2 \, a b d p \log \left (a x + b\right ) - b^{2} p e \log \left (a x + b\right ) + 2 \, a^{2} d x \log \left (c\right )}{2 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*log(c*(a+b/x)^p),x, algorithm="giac")

[Out]

1/2*(a^2*p*x^2*e*log(a*x + b) - a^2*p*x^2*e*log(x) + 2*a^2*d*p*x*log(a*x + b) + a^2*x^2*e*log(c) - 2*a^2*d*p*x
*log(x) + a*b*p*x*e + 2*a*b*d*p*log(a*x + b) - b^2*p*e*log(a*x + b) + 2*a^2*d*x*log(c))/a^2